Q. For an object thrown at $45{^\circ}$ to horizontal, the maximum height $(H)$ and horizontal range $(R)$ are related as
A) $R = 16H$
B) $R = 8H$
C) $R = 4H$
D) $R = 2H$
Explanation
(c) For $\theta = 45{^\circ}$
$H_{\max} = \frac{u^{2}\sin^{2}{45{^\circ}}}{2g} = \frac{u^{2}}{4g}\ \ \ \ \ \ \ \left\lbrack \because\sin{45{^\circ}} = \frac{1}{\sqrt{2}} \right\rbrack$
$R = \frac{u^{2}\sin{90{^\circ}}}{g} = \frac{u^{2}}{g};\ \therefore\frac{R}{H} = \frac{u^{2}}{g} \times \frac{4g}{u^{2}} = 4 \Rightarrow R = 4H$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.